depression of freezing point Questions and Answers - TopperLearning - Free Printable

Let's solve Question 11 and Question 12 step by step.

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Question 11:

> Normal boiling point of water is 373 K (at 760 mm). Vapour pressure of water at 298 K is 23 mm. If the enthalpy of evaporation is 40.656 kJ/mol, the boiling point of water at 23 mm pressure will be?

We are given:

- $ T_1 = 373 \text{ K} $, $ P_1 = 760 \text{ mm Hg} $
- $ P_2 = 23 \text{ mm Hg} $
- $ \Delta H_{\text{vap}} = 40.656 \text{ kJ/mol} = 40656 \text{ J/mol} $
- We need to find $ T_2 $, the boiling point at 23 mm Hg.

We use the Clausius-Clapeyron equation:

$$
\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)
$$

Where:
- $ R = 8.314 \text{ J/mol·K} $

Plug in the values:

$$
\ln\left(\frac{23}{760}\right) = -\frac{40656}{8.314} \left( \frac{1}{T_2} - \frac{1}{373} \right)
$$

First, compute the left-hand side:

$$
\frac{23}{760} \approx 0.030263
\quad \Rightarrow \quad \ln(0.030263) \approx -3.492
$$

Now compute the constant term:

$$
\frac{40656}{8.314} \approx 4890.2
$$

So,

$$
-3.492 = -4890.2 \left( \frac{1}{T_2} - \frac{1}{373} \right)
$$

Divide both sides by -4890.2:

$$
\frac{3.492}{4890.2} \approx 0.000714 = \frac{1}{T_2} - \frac{1}{373}
$$

Now compute $ \frac{1}{373} \approx 0.002681 $

So,

$$
\frac{1}{T_2} = 0.000714 + 0.002681 = 0.003395
$$

$$
T_2 = \frac{1}{0.003395} \approx 294.5 \text{ K}
$$

So, the boiling point at 23 mm Hg is approximately 294 K

✔ Answer: (b) 294 K

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Question 12:

> A 0.2 molal aqueous solution of a weak acid (HX) is 20% ionised. The freezing point of this solution is?
> Given: $ K_f = 1.86^\circ \text{C/m} $ for water

We are to find the freezing point depression using:

$$
\Delta T_f = i \cdot K_f \cdot m
$$

Where:
- $ m = 0.2 $ molal
- $ K_f = 1.86^\circ \text{C/m} $
- $ i $ = van't Hoff factor

Since HX is a weak acid, it partially ionizes:

$$
\text{HX} \rightleftharpoons \text{H}^+ + \text{X}^-
$$

Given: 20% ionized → $ \alpha = 0.2 $

For a weak acid dissociating into 2 ions, the van't Hoff factor is:

$$
i = 1 + \alpha = 1 + 0.2 = 1.2
$$

(Note: For weak acid HX ⇌ H⁺ + X⁻, the total number of particles after dissociation is $ 1 - \alpha + \alpha + \alpha = 1 + \alpha $.)

So,

$$
\Delta T_f = 1.2 \times 1.86 \times 0.2 = 1.2 \times 0.372 = 0.4464^\circ \text{C}
$$

So, freezing point depression ≈ $ 0.446^\circ \text{C} $

Freezing point of pure water = $ 0^\circ \text{C} $

So, freezing point of solution = $ 0 - 0.446 = -0.446^\circ \text{C} \approx -0.45^\circ \text{C} $

✔ Answer: (b) –0.45°C

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✔ Final Answers:

- Q11: (b) 294 K
- Q12: (b) –0.45°C

Let me know if you'd like a diagram or further explanation!